Objective Bayesian testing for the linear combinations of normal means

Woo Dong Lee; Sang Gil Kang; Yongku Kim

doi:10.1007/s00362-016-0831-2

Statistical Papers

February 2019, Volume 60, Issue 1, pp 147–172 | Cite as

Objective Bayesian testing for the linear combinations of normal means

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Woo Dong Lee
Sang Gil Kang
Yongku Kim

Regular Article

First Online: 24 September 2016

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Abstract

This study considers objective Bayesian testing for the linear combinations of the means of several normal populations. We propose solutions based on a Bayesian model selection procedure to this problem in which no subjective input is considered. We first construct suitable priors to test the linear combinations of means based on measuring the divergence between competing models (so-called divergence-based priors). Next, we derive the intrinsic priors for which the Bayes factors and model selection probabilities are well defined. Finally, the behavior of the Bayes factors based on the DB priors, intrinsic priors, and classical test are compared in a simulation study and an example.

Keywords

Bayes factor Divergence-based prior Intrinsic prior Linear combinations of normal means Reference prior

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Appendix 1: Proof of Theorem 1

Consider model $M_1$,

$$\begin{aligned} M_1: f_1(\mathbf{x}\vert \theta _2,\ldots ,\theta _{k+1})= & {} N\left( \mathbf{x}_1\vert {{\gamma _1\theta _{10}/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \nonumber \\&\qquad \times \prod _{i=2}^k N\left( \mathbf{x}_i \vert {{\gamma _i\theta _{10}/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\theta _i,\theta _{k+1}\right) \end{aligned}$$

(22)

and model $M_2$

$$\begin{aligned} M_2: f_2(\mathbf{x}\vert \theta _{1},\ldots ,\theta _{k+1})= & {} N\left( \mathbf{x}_1\vert {{\gamma _1\theta _1/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \nonumber \\&\qquad \times \prod _{i=2}^k N\left( \mathbf{x}_i \vert {{\gamma _i\theta _1/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\theta _i,\theta _{k+1}\right) . \end{aligned}$$

(23)

Let ${\varvec{\theta }}=\theta _1$ and ${\varvec{\nu }}=(\theta _2,\ldots ,\theta _{k+1})$. Then, Kullback–Leibler-directed divergence $KL[({\varvec{\theta }}_0,{\varvec{\nu }}):({\varvec{\theta }},{\varvec{\nu }})]$ is given by

$$\begin{aligned}&KL[({\varvec{\theta }}_0,{\varvec{\nu }}):({\varvec{\theta }},{\varvec{\nu }})] \\&= \int \log \left( {N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _1+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _1 -\theta _i,\theta _{k+1}\right) \over N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _{10}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _{10}-\theta _i,\theta _{k+1}\right) } \right) \\&\quad \quad \times N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _1+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _1 -\theta _i,\theta _{k+1}\right) d\mathbf{x}\\&={1\over \sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over 2\theta _{k+1}}, \end{aligned}$$

where $g(\gamma _i)={{\gamma _i/n_i} \over {\sum _{j=1}^k\gamma _j^2/n_j}}$ and $n=n_1+\ldots +n_k$. Moreover, Kullback–Leibler-directed divergence $KL[({\varvec{\theta }},{\varvec{\nu }}):({\varvec{\theta }}_0,{\varvec{\nu }})]$ is given by

$$\begin{aligned}&KL[({\varvec{\theta }},{\varvec{\nu }}):({\varvec{\theta }}_0,{\varvec{\nu }})] \\&= \int \log \left( {N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _{10}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _{10} -\theta _i,\theta _{k+1}\right) \over N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _{1}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _1-\theta _i,\theta _{k+1}\right) } \right) \\&\quad \quad \times N\left( \mathbf{x}_1\vert g(\gamma _1)\theta _{10}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert g(\gamma _i)\theta _{10} -\theta _i,\theta _{k+1}\right) d\mathbf{x}\\&={1\over \sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over 2\theta _{k+1}}. \end{aligned}$$

Therefore, the sum divergence measure is

$$\begin{aligned} D^S[({\varvec{\theta }},{\varvec{\theta }}_0)\vert {\varvec{\nu }})] = {1\over \sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over \theta _{k+1}}. \end{aligned}$$

Further, since $KL[({\varvec{\theta }}_0,{\varvec{\nu }}):({\varvec{\theta }},{\varvec{\nu }})]$ and $KL[({\varvec{\theta }},{\varvec{\nu }}):({\varvec{\theta }}_0,{\varvec{\nu }})]$ are the same, the minimum divergence measure is the same as the sum divergence measure. We take the effective sample size $n^*=n$. Then, the unitary symmetrized divergence is

$$\begin{aligned} \bar{D}^S[({\varvec{\theta }},{\varvec{\theta }}_0)\vert {\varvec{\nu }})]= {1\over n\sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over \theta _{k+1}}. \end{aligned}$$

Now,

$$\begin{aligned} c_S(q,{\varvec{\nu }})= & {} \int (1+\bar{D}^S[({\varvec{\theta }},{\varvec{\theta }}_0) \vert {\varvec{\nu }}])^{-q}\pi ^N({\varvec{\theta }}\vert {\varvec{\nu }})d{\varvec{\theta }}\\= & {} \int _{0}^{\infty }\left[ 1+ {1\over n\sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over \theta _{k+1}} \right] ^{-q}d\theta _1\\= & {} {\Gamma (q-{1\over 2})\Gamma ({1\over 2})\over \Gamma (q)}\theta _{k+1}^{1\over 2} \left( n\sum _{j=1}^k\gamma _j^2/n_j\right) ^{1\over 2}<\infty , \end{aligned}$$

if $q> {1\over 2}$. Thus, the conditional sum-DB prior with $q_*^S=1$ is given by

$$\begin{aligned}&\pi ^S(\theta _1\vert \theta _2,\ldots ,\theta _{k+1}) = \pi ^{-1}\theta _{k+1}^{-{1\over 2}}\left( n\sum _{j=1}^k\gamma _j^2/n_j\right) ^{-{1\over 2}}\nonumber \\&\quad \left[ 1+ {1\over n\sum _{j=1}^k\gamma _j^2/n_j}{(\theta _1-\theta _{10})^2\over \theta _{k+1}} \right] ^{-1}. \end{aligned}$$

This proves Theorem 1.

Appendix 2: Proof of Theorem 2

The likelihood function under model $M_1$ is

$$\begin{aligned} L_1(\theta _2,\ldots ,\theta _{k+1}\vert \mathbf {x} ) = \left[ \prod _{i=1}^k \left( {1\over 2\pi \theta _{k+1}}\right) ^{n_i\over 2}\right] \exp \left\{ -\sum _{i=1}^k {1\over 2\theta _{k+1}} (S_i^2+T_{1i}^2) \right\} , \end{aligned}$$

(24)

where $Y_{i\cdot }=\sum _{j=1}^{n_i} Y_{ij} /n_i$, $S_i^2=\sum _{j=1}^{n_i}(Y_{ij}-Y_{i\cdot })^2, i=1,\ldots ,k,$$T_{11}^2=n_1(Y_{1\cdot }- {{\gamma _1\theta _{10}/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}} - \sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i)^2$, and $T_{1i}^2=n_i(Y_{i\cdot }- {{\gamma _i\theta _{10}/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} +\theta _i)^2, i=2,\ldots ,k$. In addition, under model $M_1$, the reference prior for $(\theta _2,\ldots ,\theta _{k+1})$ is

$$\begin{aligned} \pi _1^N(\theta _2,\ldots ,\theta _{k+1}) \propto \theta _{k+1}^{-1}. \end{aligned}$$

(25)

Then, from the likelihood (24) and reference prior (25), $m_1(\mathbf {x})$ is given by

Open image in new window

where $n=n_1+\cdot +n_k$, $c_1=n_1$, $c_i=[c_{i-1}{\gamma _i^2\over \gamma _1^2}+n_i]^{-1}c_{i-1}n_i, i\ge 2$. For model $M_2$, the reference prior for $(\theta _1,\ldots ,\theta _{k+1})$ is

$$\begin{aligned} \pi ^N (\theta _1,\ldots ,\theta _{k+1}) \propto \theta _{k+1}^{-1}. \end{aligned}$$

(26)

The likelihood function under model $M_2$ is

$$\begin{aligned} L_2(\theta _1,\ldots ,\theta _{k+1}\vert \mathbf {x} ) = \left[ \prod _{i=1}^k \left( {1\over 2\pi \theta _{k+1}}\right) ^{n_i\over 2}\right] \exp \left\{ -\sum _{i=1}^k {1\over 2\theta _{k+1}} (S_i^2+T_{2i}^2) \right\} , \end{aligned}$$

(27)

where $T_{21}^2=n_1(Y_{1\cdot }-{{\gamma _1\theta _{1}/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}} - \sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i)^2$ and $T_{2i}^2=n_i(Y_{i\cdot }- {{\gamma _i\theta _{1}/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} +\theta _i)^2, i=2,\ldots ,k$. Let

$$\begin{aligned} \mu _1= {{(\gamma _1/n_1)}\theta _1 \over {\sum _{j=1}^k \gamma _j^2/n_j}} +\sum _{i=2}^k {\gamma _i \over \gamma _1}\theta _i, ~\mu _i = {{(\gamma _i/n_i)}\theta _1 \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\theta _i, i=2,\ldots ,k, ~\sigma ^2=\theta _{k+1}. \end{aligned}$$

Then, from the likelihood (27) and reference prior (26), $m_2(\mathbf {x})$ is given as follows:

Open image in new window

Therefore, $B_{21}^N$ is given by

$$\begin{aligned} B_{21} ^{N}={g_2(\mathbf{x})\over g_1(\mathbf{x})}, \end{aligned}$$

(28)

where

$$\begin{aligned} g_2(\mathbf{x})=(2\pi )^{-{n-k\over 2}}\left[ \prod _{i=1}^{k}n_i^{-{1\over 2}}\right] \Gamma \left( {n-k\over 2}\right) \left[ {\sum _{i=1}^k S_i^2 \over 2}\right] ^{-{n-k\over 2}}|\gamma _1| \end{aligned}$$

and

$$\begin{aligned} g_1(\mathbf{x})= & {} (2\pi )^{-{n-k+1\over 2}}\left[ \prod _{i=2}^{k} \left( c_{i-1}{\gamma _i^2\over \gamma _1^2}+n_i\right) ^{-{1\over 2}}\right] \Gamma \left( {n-k+1\over 2}\right) \\&\quad \times \left[ {\sum _{i=1}^k S_i^2 +\gamma _1^{-2}c_k(\sum _{i=1}^k \gamma _i\bar{y}_i-\theta _{10})^2\over 2}\right] ^{-{n-k+1\over 2}}. \end{aligned}$$

Further, $\pi ^N(\theta _1\vert \theta _2,\ldots ,\theta _{k+1})=1$. Hence, Theorem 2 is proven. $\square $

Appendix 3: Proof of Lemma 1

Consider the models $M_1$,

$$\begin{aligned} M_1: f_1(\mathbf{x}\vert {\varvec{\theta }}_1)= & {} N\left( \mathbf{x}_1\vert {{\gamma _1\theta _{10}/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\tau _i,\tau _{k+1}\right) \\&\quad \times \prod _{i=2}^k N\left( \mathbf{x}_i \vert {{\gamma _i\theta _{10}/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\tau _i,\tau _{k+1}\right) , \pi ^N(\tau _2,\ldots ,\tau _{k+1})=c_1\tau _{k+1}^{-1} \end{aligned}$$

and $M_2$,

$$\begin{aligned} M_2: f_2(\mathbf{x}\vert {\varvec{\theta }}_2)= & {} N\left( \mathbf{x}_1\vert {{\gamma _1\theta _1/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \\&\quad \quad \times \prod _{i=2}^k N\left( \mathbf{x}_i \vert {{\gamma _i\theta _1/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} {-}\theta _i,\theta _{k{+}1}\right) , \pi ^N(\theta _{1},\ldots ,\theta _{k+1})=c_2\theta _{k+1}^{-1}, \end{aligned}$$

where $c_1$ and $c_2$ are arbitrary positive constants. For the minimal training sample $z_i(l)$, we have

$$\begin{aligned} B_{12}^N (z_i(l))= & {} {1\over m_2^N (z_i(l))} \left[ N\left( x_{1j}\vert g(\gamma _1)\theta _{10}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\tau _i,\tau _{k+1}\right) \right. \\&\quad \quad \times \left. \prod _{j=1}^{2} N\left( x_{ij}\vert g(\gamma _i)\theta _{10} -\tau _i,\tau _{k+1}\right) \times \prod _{l\ne i} N\left( x_{l1}\vert g(\gamma _l)\theta _{10} -\tau _l,\tau _{k+1}\right) \right] , \end{aligned}$$

where $g(\gamma _i)={{\gamma _i/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}}$ and $ m_2^N (z_i(l))={c_2 |\gamma _1|\over |x_{i1}-x_{i2}|}$. Let $u=x_{i1}-x_{i2}$ and $v=x_{i1}+x_{i2}$. Then, by direct integration on $x_{l1}$ and (v, u), we obtain

$$\begin{aligned}&E_{z_i(l)\vert {\varvec{\theta }}_2}^{M_2} [ B_{12}^N (z_i(l))]\\&= \int {1\over 2\pi }\tau _{k+1}^{-{1\over 2}}\theta _{k+1}^{-{1\over 2}} \exp \left\{ -{(x_{11}-g(\gamma _1)\theta _{10}-\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\tau _i)^2\over 2\tau _{k+1}} {-}{(x_{11}-g(\gamma _1)\theta _{1} {-}\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i)^2\over 2\theta _{k+1}} \right\} dx_{11}\\&\quad \times \int \int {1\over c_2|\gamma _1|}{1\over (2\pi )^2 }\tau _{k+1}^{-1}\theta _{k+1}^{-1} |x_{i1}-x_{i2}| \exp \left\{ -{(x_{i1}-g(\gamma _i)\theta _{10} +\tau _i)^2+(x_{i2}-g(\gamma _i)\theta _{10} +\tau _i)^2\over 2\tau _{k+1}}\right. \\&\left. \quad -{(x_{i1}-g(\gamma _i)\theta _{1} +\theta _i)^2+(x_{i2}-g(\gamma _i)\theta _{1}+\theta _i)^2\over 2\theta _{k+1}}\right\} dx_{i1}dx_{i2}\\&\quad \times \prod _{l\ne i}\int {1\over 2\pi }\tau _{k+1}^{-{1\over 2}}\theta _{k+1}^{-{1\over 2}} \exp \left\{ -{(x_{l1}-g(\gamma _l)\theta _{10} +\tau _l)^2\over 2\tau _{k+1}} -{(x_{l1}-g(\gamma _l)\theta _{1} +\theta _l)^2\over 2\theta _{k+1}} \right\} dx_{l1}\\&={1\over c_2|\gamma _1|}{1\over \pi }{\tau _{k+1}^{1\over 2}\theta _{k+1}^{1\over 2}\over (\theta _{k+1}+\tau _{k+1})}\\&\quad \times {1\over \sqrt{2\pi }(\tau _{k+1}+\theta _{k+1})^{1\over 2}} \exp \left\{ {(g(\gamma _1)\theta _{1}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i -g(\gamma _1)\theta _{10}-\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\tau _i)^2\over 2(\tau _{k+1}+\theta _{k+1})}\right\} \\&\quad \times {1\over \sqrt{\pi }(\tau _{k+1}+\theta _{k+1})^{1\over 2}} \exp \left\{ -{(\theta _i-g(\gamma _i)\theta _{1} -\tau _i+g(\gamma _i)\theta _{10})^2\over \tau _{k+1}+\theta _{k+1}}\right\} \\&\quad \times \prod _{l\ne i}{1\over \sqrt{2\pi }(\tau _{k+1}+\theta _{k+1})^{1\over 2}} \exp \left\{ -{(\theta _l-g(\gamma _l)\theta _{1} -\tau _l+g(\gamma _l)\theta _{10})^2\over 2(\tau _{k+1}+\theta _{k+1})}\right\} . \end{aligned}$$

Therefore,

Open image in new window

Finally, we can similarly derive the intrinsic prior for the other minimal training sample. This proves Lemma 1. $\square $

Appendix 4: Proof of Theorem 4

We compute the Bayes factor to compare model $M_1$ and model $M_2$ with the intrinsic prior $\pi _{2i}^I(\theta _2)$. We easily obtain the marginal density $m_1(\mathbf{x})$ from Theorem 2. Hence, we only compute the marginal density $m_2(\mathbf{x})$. Now,

$$\begin{aligned}&f_2(\mathbf {x} \vert {\varvec{\theta }}_2) \pi _{2i}^I({\varvec{\theta }}_2\vert {\varvec{\theta }}_1)\pi _1^N({\varvec{\theta }}_1)= {c_1\over \pi |\gamma _1|}{\tau _{k+1}^{-{1\over 2}}\over \theta _{k+1}^{1\over 2}(\theta _{k+1}+\tau _{k+1})}\nonumber \\&\quad \times \, N\left( \mathbf{x}_1\vert {{\gamma _1\theta _1/n_1} \over {\sum _{j=1}^k \gamma _j^2/n_j}}+\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i,\theta _{k+1}\right) \prod _{i=2}^k N\left( \mathbf{x}_i \vert {{\gamma _i\theta _1/n_i} \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\theta _i,\theta _{k+1}\right) \nonumber \\&\quad \times \, N\left( {{\gamma _1\theta _1/n_1} \over {\sum _{j=1}^k\gamma _j^2/n_j}}\vert {{\gamma _1\theta _{10}/n_1}\over {\sum _{j=1}^k\gamma _j^2/n_j}} +\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\tau _i-\sum _{i=2}^{k}{\gamma _i\over \gamma _1}\theta _i, \theta _{k+1}+\tau _{k+1}\right) \nonumber \\&\quad \times \, N\left( \theta _{i}\vert \tau _i-{{\gamma _i\theta _{10}/n_i}\over {\sum _{j=1}^k\gamma _j^2/n_j}} +{{\gamma _i\theta _1/n_i}\over {\sum _{j=1}^k\gamma _j^2/n_j}}, {\theta _{k+1}+\tau _{k+1}\over 2}\right) \nonumber \\&\quad \times \, \prod _{l\ne i\ge 2}N\left( \theta _{l}\vert \tau _l-{{\gamma _l\theta _{10}/n_l}\over {\sum _{j=1}^k\gamma _j^2/n_j}} +{{\gamma _l\theta _1/n_l}\over {\sum _{j=1}^k\gamma _j^2/n_j}}, {\theta _{k+1}+\tau _{k+1}}\right) . \end{aligned}$$

(29)

Let

$$\begin{aligned} \mu _1= {{(\gamma _1/n_1)}\theta _1 \over {\sum _{j=1}^k \gamma _j^2/n_j}} +\sum _{i=2}^k {\gamma _i \over \gamma _1}\theta _i, ~\mu _i = {{(\gamma _i/n_i)}\theta _1 \over {\sum _{j=1}^k \gamma _j^2/n_j}} -\theta _i, i=2,\ldots ,k. \end{aligned}$$

By integrating with respect to $\mu _i, i=1,\ldots ,k$, and $\tau _i, i=2,\ldots ,k$ in (29), we get

$$\begin{aligned}&\pi ^{-1}(2\pi )^{-{n-k+1\over 2}}2^{-{k-2\over 2}} \left[ \prod _{i=2}^{k}\left( p_{i-1}{\gamma _i^2\over \gamma _1^2}+d_i\right) ^{-{1\over 2}}\right] \theta _{k+1}^{-{n-k+1\over 2}}{\tau _{k+1}^{-{1\over 2}}\over \theta _{k+1}^{1\over 2}(\theta _{k+1}+\tau _{k+1})}\nonumber \\&\quad \times \left[ n_i\left( 1+{\tau _{k+1}\over \theta _{k+1}}\right) +2\right] ^{-{1\over 2}} \prod _{j\ne i}\left[ n_j\left( 1+{\tau _{k+1}\over \theta _{k+1}}\right) +1\right] ^{-{1\over 2}}\nonumber \\&\quad \times \exp \left\{ -{1\over \theta _{k+1}}\left[ {\sum _{i=1}^k s_i^2\over 2} +\gamma _1^{-2}p_k\left( \sum _{i=1}^k \gamma _i\bar{y}_i-\theta _{10}\right) ^2\right] \right\} , \end{aligned}$$

(30)

where $p_1={n_1\over 2n_1(1+{\tau _{k+1}\over \theta _{k+1}})+2}$, $p_i=[p_{i-1}{\gamma _i^2\over \gamma _1^2}+d_i]^{-1}p_{i-1}d_i$, $d_i={n_i\over n_i(1+{\tau _{k+1}\over \theta _{k+1}})+2}$, $d_{j\ne i}={n_j\over 2n_j(1+{\tau _{k+1}\over \theta _{k+1}})+2}$. Let $\theta _{k+1}=\theta _{k+1}$ and $\omega =\tau _{k+1}/\theta _{k+1}$. By integrating with respect to $\theta _{k+1}$ in (30), we get

$$\begin{aligned} m_2(\mathbf x)= & {} \int _{0}^{\infty } \pi ^{-1}(2\pi )^{-{n-k+1\over 2}}2^{-{k-2\over 2}}\Gamma \left( {n-k+1\over 2}\right) \left[ \prod _{i=2}^{k}\left( p_{i-1}{\gamma _i^2\over \gamma _1^2}+d_i\right) ^{-{1\over 2}}\right] \nonumber \\&\quad \times \left[ {\sum _{i=1}^k S_i^2\over 2} +\gamma _1^{-2}p_k(\sum _{i=1}^k \gamma _i\bar{y}_i-\theta _{10})^2\right] ^{-{n-k+1\over 2}}\nonumber \\&\quad \times (1+\omega )^{-1}\omega ^{-{1\over 2}}[n_i(1+\omega )+2]^{-{1\over 2}} \prod _{j\ne i}[n_j(1+\omega )+1]^{-{1\over 2}}d\omega . \end{aligned}$$

(31)

Finally, we can similarly compute the marginal density by using other intrinsic priors. Hence, Theorem 4 is proven.

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Woo Dong Lee
- 1
Sang Gil Kang
- 2
Yongku Kim
- 3
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1.Department of Data ManagementDaegu Haany UniversityKyungsanKorea
2.Department of Computer and Data InformationSangji UniversityWonjuKorea
3.Department of StatisticsKyungpook National UniversityDaeguKorea

Objective Bayesian testing for the linear combinations of normal means

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